Đáp án:
a, `|2x - 3| - 1 = 0`
`|2x - 3| = 1`
⇒\(\left[ \begin{array}{l}2x - 3 = 1\\2x - 3 = -1\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x = 2\\x = 1\end{array} \right.\)
Vậy `x \in {2 ; 1}`
b, `2 . |3x - 1/2| + 5 = 6`
`2 . | 3x - 1/2| = 1`
` |3x - 1/2| = 1/2`
⇒\(\left[ \begin{array}{l}3x - \frac{1}{2} = \frac{1}{2}\\3x - \frac{1}{2} = -\frac{1}{2}\end{array} \right.\)
⇒\(\left[ \begin{array}{l}3x = 1\\3x = 0\end{array} \right.\)
⇒\(\left[ \begin{array}{l}x = \frac{1}{3}\\x = 0\end{array} \right.\)
Vậy `x \in{ 1/3 ; 0}`
